3.292 \(\int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx\)

Optimal. Leaf size=113 \[ -\frac {1}{9} \left (\frac {1-x^3}{x^3+1}\right )^{3/2} \left (x^3+1\right )^3-\frac {1}{6} \sqrt {\frac {1-x^3}{x^3+1}} \left (x^3+1\right )^2+\frac {1}{2} \sqrt {\frac {1-x^3}{x^3+1}} \left (x^3+1\right )-\frac {1}{3} \tan ^{-1}\left (\sqrt {\frac {1-x^3}{x^3+1}}\right ) \]

[Out]

-1/9*((-x^3+1)/(x^3+1))^(3/2)*(x^3+1)^3-1/3*arctan(((-x^3+1)/(x^3+1))^(1/2))+1/2*(x^3+1)*((-x^3+1)/(x^3+1))^(1
/2)-1/6*(x^3+1)^2*((-x^3+1)/(x^3+1))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1960, 463, 455, 385, 204} \[ -\frac {1}{9} \left (\frac {1-x^3}{x^3+1}\right )^{3/2} \left (x^3+1\right )^3-\frac {1}{6} \sqrt {\frac {1-x^3}{x^3+1}} \left (x^3+1\right )^2+\frac {1}{2} \sqrt {\frac {1-x^3}{x^3+1}} \left (x^3+1\right )-\frac {1}{3} \tan ^{-1}\left (\sqrt {\frac {1-x^3}{x^3+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^8*Sqrt[(1 - x^3)/(1 + x^3)],x]

[Out]

(Sqrt[(1 - x^3)/(1 + x^3)]*(1 + x^3))/2 - (Sqrt[(1 - x^3)/(1 + x^3)]*(1 + x^3)^2)/6 - (((1 - x^3)/(1 + x^3))^(
3/2)*(1 + x^3)^3)/9 - ArcTan[Sqrt[(1 - x^3)/(1 + x^3)]]/3

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

Rule 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx &=-\left (\frac {4}{3} \operatorname {Subst}\left (\int \frac {x^2 \left (-1+x^2\right )^2}{\left (-1-x^2\right )^4} \, dx,x,\sqrt {\frac {1-x^3}{1+x^3}}\right )\right )\\ &=-\frac {1}{9} \left (\frac {1-x^3}{1+x^3}\right )^{3/2} \left (1+x^3\right )^3-\frac {2}{9} \operatorname {Subst}\left (\int \frac {x^2 \left (6-6 x^2\right )}{\left (-1-x^2\right )^3} \, dx,x,\sqrt {\frac {1-x^3}{1+x^3}}\right )\\ &=-\frac {1}{6} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )^2-\frac {1}{9} \left (\frac {1-x^3}{1+x^3}\right )^{3/2} \left (1+x^3\right )^3+\frac {1}{18} \operatorname {Subst}\left (\int \frac {12-24 x^2}{\left (-1-x^2\right )^2} \, dx,x,\sqrt {\frac {1-x^3}{1+x^3}}\right )\\ &=\frac {1}{2} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )-\frac {1}{6} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )^2-\frac {1}{9} \left (\frac {1-x^3}{1+x^3}\right )^{3/2} \left (1+x^3\right )^3+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {\frac {1-x^3}{1+x^3}}\right )\\ &=\frac {1}{2} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )-\frac {1}{6} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )^2-\frac {1}{9} \left (\frac {1-x^3}{1+x^3}\right )^{3/2} \left (1+x^3\right )^3-\frac {1}{3} \tan ^{-1}\left (\sqrt {\frac {1-x^3}{1+x^3}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 98, normalized size = 0.87 \[ \frac {\sqrt {\frac {1-x^3}{x^3+1}} \sqrt {x^3+1} \left (6 \sqrt {1-x^3} \sin ^{-1}\left (\frac {\sqrt {1-x^3}}{\sqrt {2}}\right )+\sqrt {x^3+1} \left (2 x^9-5 x^6+7 x^3-4\right )\right )}{18 \left (x^3-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8*Sqrt[(1 - x^3)/(1 + x^3)],x]

[Out]

(Sqrt[(1 - x^3)/(1 + x^3)]*Sqrt[1 + x^3]*(Sqrt[1 + x^3]*(-4 + 7*x^3 - 5*x^6 + 2*x^9) + 6*Sqrt[1 - x^3]*ArcSin[
Sqrt[1 - x^3]/Sqrt[2]]))/(18*(-1 + x^3))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 65, normalized size = 0.58 \[ \frac {1}{18} \, {\left (2 \, x^{9} - x^{6} + x^{3} + 4\right )} \sqrt {-\frac {x^{3} - 1}{x^{3} + 1}} - \frac {1}{3} \, \arctan \left (\frac {{\left (x^{3} + 1\right )} \sqrt {-\frac {x^{3} - 1}{x^{3} + 1}} - 1}{x^{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*((-x^3+1)/(x^3+1))^(1/2),x, algorithm="fricas")

[Out]

1/18*(2*x^9 - x^6 + x^3 + 4)*sqrt(-(x^3 - 1)/(x^3 + 1)) - 1/3*arctan(((x^3 + 1)*sqrt(-(x^3 - 1)/(x^3 + 1)) - 1
)/x^3)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{8} \sqrt {-\frac {x^{3} - 1}{x^{3} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*((-x^3+1)/(x^3+1))^(1/2),x, algorithm="giac")

[Out]

integrate(x^8*sqrt(-(x^3 - 1)/(x^3 + 1)), x)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 80, normalized size = 0.71 \[ -\frac {\sqrt {-\frac {x^{3}-1}{x^{3}+1}}\, \sqrt {-\left (x^{3}+1\right ) \left (x^{3}-1\right )}\, \arcsin \left (x^{3}\right )}{6 \left (x^{3}-1\right )}+\frac {\left (2 x^{6}-3 x^{3}+4\right ) \left (x^{3}+1\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}}{18} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*((-x^3+1)/(x^3+1))^(1/2),x)

[Out]

1/18*(2*x^6-3*x^3+4)*(x^3+1)*(-(x^3-1)/(x^3+1))^(1/2)-1/6*(-(x^3-1)/(x^3+1))^(1/2)*(-(x^3+1)*(x^3-1))^(1/2)/(x
^3-1)*arcsin(x^3)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{8} \sqrt {-\frac {x^{3} - 1}{x^{3} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*((-x^3+1)/(x^3+1))^(1/2),x, algorithm="maxima")

[Out]

integrate(x^8*sqrt(-(x^3 - 1)/(x^3 + 1)), x)

________________________________________________________________________________________

mupad [B]  time = 2.66, size = 101, normalized size = 0.89 \[ \frac {2\,\sqrt {-\frac {x^3-1}{x^3+1}}}{9}-\frac {\mathrm {atan}\left (\sqrt {-\frac {x^3-1}{x^3+1}}\right )}{3}+\frac {x^3\,\sqrt {-\frac {x^3-1}{x^3+1}}}{18}-\frac {x^6\,\sqrt {-\frac {x^3-1}{x^3+1}}}{18}+\frac {x^9\,\sqrt {-\frac {x^3-1}{x^3+1}}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(-(x^3 - 1)/(x^3 + 1))^(1/2),x)

[Out]

(2*(-(x^3 - 1)/(x^3 + 1))^(1/2))/9 - atan((-(x^3 - 1)/(x^3 + 1))^(1/2))/3 + (x^3*(-(x^3 - 1)/(x^3 + 1))^(1/2))
/18 - (x^6*(-(x^3 - 1)/(x^3 + 1))^(1/2))/18 + (x^9*(-(x^3 - 1)/(x^3 + 1))^(1/2))/9

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*((-x**3+1)/(x**3+1))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________